###### Calculating Unconditional Probability ...

We can use the total probability rule to determine the unconditional probability of... **Read More**

Counting problems involve determination of the exact number of ways two or more operations or events can be performed together. For instance, we might be interested in the number of ways to choose 7 chartered analysts comprising 3 women and 4 men from a group of 50 analysts. Counting encompasses the following fundamental principles:

“*n* factorial” (*n!*) is used to represent the product of the first *n ***natural numbers. **Generally:

$$ n! = n * (n – 1) * (n – 2) * (n – 3) * … * 2 * 1 $$

For example,

1! = 1 * 1

2! = 2 * 1

3! = 3 * 2 * 1

4! = 4 * 3 * 2 * 1

*Note to candidates: 0! is just 1, not zero. You should also remember that we can find n! only if *n* is a whole number. Thus, we cannot have 1.8!*

The labeling principle is used to assign k labels or groups to a total of *n* items, where each label contains n_{i} items such that n_{1 }+ n_{2 }+ n_{3} + … + n_{k }= n. In other words, your wish is to have *n* items categorized into k groups, where the number of items in each group is **pre-determined**. To get the total number of ways that the labels or groups can be assigned, you use the formula:

$$ \cfrac { n! }{ { n }_{ 1 }*{ n }_{ 2 }*{ n }_{ 3 }*…*{ n }_{ k } }

$$

**Example 1**

Assume that you have a portfolio of investments consisting of 10 stocks. Suppose your wish is to assign 3 different labels such that label 1 has 5 “high return” stocks, label 2 has 3 “medium return” stocks, and the last label has 2 “low return” stocks.

**Solution**

n = 10

There are 3 labels, where n_{1} = 5, n_{2} = 3, and n_{3 }= 2.

The number of different ways that you can assign the 3 labels =\(\cfrac {10!}{(5! * 3! * 2!)} = 2520 \text{ ways}.\)

A combination is basically a selection of some given items where the order **does not** matter. The number of combinations (possible ways) of *n* items taken *r* at a time is:

$$ n{ C }r=\cfrac { n! }{ \left( n-r \right) !r! } $$

**Example 2**

In how many ways can we choose 3 stocks from a portfolio consisting of 10 stocks?

**Solution**

We are interested in the number of combinations of 10 items, taken 3 at a time. Therefore,

n = 10

r = 3

Number of possible combinations =\(\cfrac {10!}{(7! * 3!)} = 120 \)

Unlike a combination, a permutation involves determination of the number of possible ways to choose *r* items from *n* items **where the order is paramount**. Simply put**, **the order of the *r* items chosen matters, i.e., which one comes first? Which one should come last?

$$ \text{The number of Permutations}, \text{nPr}=\cfrac { n! }{ \left( n-r \right) ! } $$

**Example 3**

Let us revisit example 2. Imagine the three chosen stocks are to be sold, in an arrangement where the order of sale is important.

**Solution**

This means that once we have chosen 3 stocks, we must also determine the order in which to sell them. Thus, the number of possible permutations = 10!/7! = 720.

*Note to Candidates: if you compare the combination formula and the permutation formula, the only difference is the r! in the denominator of the former. This means that in any situation, there are always r! more ways to choose items when the order is important compared to when the order is not important. For instance, note that 720 is just 3! multiplied by 120.*

**How Do You Determine the Approach to Take?**

- If you are asked to assign
*n*items to*n*slots, use the factorial formula. - In case you are asked to assign k unique labels or categories to
*n*items, use the labeling formula. - When asked to come up with a number of ways to choose
*r*items from*n*items when the order is not important, use the combination formula. If the order is important, use the permutation formula.

*Reading 8 LOS 8o:*

*Identify the most appropriate method to solve a particular counting problem and solve counting problems using factorial, combination, and permutation concepts.*